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1)For three events A, B and C, we know that

• A and C are independent
B and C are independent
A and B are disjoint
P(A∪C)=2/3 P(B∪C)=3/4 P(A∪B∪C)=11/12

P(A)=___________  ans  1/3

2)Consider independent trails consisting of rolling a pair of fair dice, over and over. What is the probability that a sum of 5 appears before sum of 7? ans 2/5

Answer to question no 1 :

Given A and B are disjoint , so P( A ∩  B )  = 0

Given B and C are independent  =  P(B ∩  C)  =  P(B) . P(C)

A and C are independent  = P(A ∩  C)  = P(A) . P(C)

As A and B are disjoint , then A , B and C will obviosuly be disjoint which is the trick of the question.

We know ,

P(A U B U C) = P(A) + P(B) + P(C) - P( A ∩  B ) - P(B ∩  C) - P(A  ∩  C)  +  P(A ∩  B   ∩  C)

==>     11 / 12        =  x + y + z  - yz - xz   [ Say ]   ..............(1)

P(B U C)      =   3 /4

==>    P(B) + P(C) - P(B ∩  C)  =  3 /4

==>     y + z  - yz    =   3 / 4      .....(2)

Substituting in (1) , we have :

==>   11 / 12     =   x +  ( 3 / 4 )  -  xz  .............(3)

Also   P(A U C)  =  2 / 3

==>   P(A) + P(C) - P(A).P(C)  =  2 / 3

==>   x  + z  -  xz   = 2 / 3

==>   x  -  xz   =  2/3 - z

So substituting this in (3) , we have :

11 / 12  =  (2 / 3   -  z)  +  (3 / 4)

==>  z         =  (2 / 3 + 3 / 4  - 11 / 12)

==>  z         =   6 / 12

Now x - xz  =  2/3 - z

==> x - 1/2x  = 2/3 - 1 /2

==> 1/2 x    = 1 / 6

==>  x       =  1 / 3

Therefore P(A)  =  1 / 3

You are interested that the game will end where you first get sum of 5, and that it will happen before the first "sum if 7". Hence, by noting that the first event has 4 elementary outcomes(i.e. of sum of 5) while the second has 6(i.e. of sum of 7)..

So this can be done by in the first "n-1" trials , the remaining 26 ( 36 - 4 - 6 ) outcomes can come and in the nth trial , we need to ensure that only these 4 outcomes which constitute sum of 5 comes..That way we can ensure that sum of 5 will come definitely before sum of 7..So an infinite G.P. will perform as n can be any natural number ranging from 1 to infinity [In the very 1st trial we can get sum of 5 and then we are done..But we need to consider all cases m so infinite G.P will form as ] :

P(5 comes before 7 as sum in 2 dices)  =  (4 / 36) + (26 / 36) * (4 / 36) + (26 / 36)2 * (4 / 36) ..............to infinity

=  (4 /  36) [ 1 + (26 / 36) + (26 / 36)2 + .................. ]

=  ( 1 / 9 ) * [1 / ( 1  -  (26 / 36) }

=    1 / 9 * (36 / 10)

=    4 / 10

=    2 / 5

Hence 2 / 5 is required probability here..